Let AD and BC be two vertical poles at A and | vedclass

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(A) Let the point $A$ be at the origin $(0, 0)$. Since $AB = 10 \ m$,the coordinates are $A(0, 0)$ and $B(10, 0)$.The poles are vertical,so $D$ is at

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$5$

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(A) Let the point $A$ be at the origin $(0, 0)$. Since $AB = 10 \ m$,the coordinates are $A(0, 0)$ and $B(10, 0)$.The poles are vertical,so $D$ is at $(0, 8)$ and $C$ is at $(10, 11)$.Let $M$ be a point on $AB$ at a distance $h$ from $A$,so $M$ has coordinates $(h, 0)$ where $0 \le h \le 10$.The squared distances are $MD^{2} = (h-0)^{2} + (0-8)^{2} = h^{2} + 64$ and $MC^{2} = (h-10)^{2} + (0-11)^{2} = (h-10)^{2} + 121$.Let $f(h) = MD^{2} + MC^{2} = h^{2} + 64 + (h-10)^{2} + 121$.$f(h) = h^{2} + 64 + h^{2} - 20h + 100 + 121 = 2h^{2} - 20h + 285$.To find the minimum,we take the derivative $f'(h) = 4h - 20$.Setting $f'(h) = 0$ gives $4h = 20$,so $h = 5$.Since $f''(h) = 4 > 0$,the function is minimum at $h = 5 \ m$.

Let $AD$ and $BC$ be two vertical poles at $A$ and $B$ respectively on a horizontal ground. If $AD = 8 \ m$,$BC = 11 \ m$ and $AB = 10 \ m$; then the distance (in meters) of a point $M$ on $AB$ from the point $A$ such that $MD^{2} + MC^{2}$ is minimum is

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